\newproblem{lay:5_5_26}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 5.5.26}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Jan. 20th, 2014} \\}{}

  % Problem statement
	Let $A$ be a real $2\times 2$ matrix with a complex eigenvalue $\lambda=a-bi$ ($b\neq 0$) and an associated eigenvector in $\mathbf{v}\in\mathbb{C}^2$.
	\begin{enumerate}
		\item Show that $A\mathrm{Real}\{\mathbf{v}\}=a\mathrm{Real}\{\mathbf{v}\}+b\mathrm{Imag}\{\mathbf{v}\}$ and 
			    $A\mathrm{Imag}\{\mathbf{v}\}=-b\mathrm{Real}\{\mathbf{v}\}+a\mathrm{Imag}\{\mathbf{v}\}$ (\textit{Hint:} Write
					$\mathbf{v}=\mathrm{Real}\{\mathbf{v}\}+i\mathrm{Imag}\{\mathbf{v}\}$ and compute $A\mathbf{v}$.)
		\item Verify that if $A$ is diagonalized as $A=PCP^{-1}$, then $AP=PC$.
	\end{enumerate}
}{
  % Solution
	\begin{enumerate}
		\item Let us calculate $A\mathbf{v}$
		      \begin{equation}
						\begin{array}{rcl}
							A\mathbf{v}&=&\lambda\mathbf{v} \\
							           &=&(a-bi)\mathrm{Real}\{\mathbf{v}\}+i\mathrm{Imag}\{\mathbf{v}\} \\
												 &=&a\mathrm{Real}\{\mathbf{v}\}+ia\mathrm{Imag}\{\mathbf{v}\}-bi\mathrm{Real}\{\mathbf{v}\}+b\mathrm{Imag}\{\mathbf{v}\}\\
												 &=&(a\mathrm{Real}\{\mathbf{v}\}+b\mathrm{Imag}\{\mathbf{v}\})+i(a\mathrm{Imag}\{\mathbf{v}\}-b\mathrm{Real}\{\mathbf{v}\})
						\end{array}
					\end{equation}
		\item It suffices to multiply by $P$ on the right to get
					\begin{equation}
						AP=(PCP^{-1})P=PC(P^{-1}P)=PC
					\end{equation}
	\end{enumerate}
}
\useproblem{lay:5_5_26}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
